Question 1205772
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Answer: <font color=red size=4>(2x+3)</font> which is <font color=red size=4>choice B</font>


Explanation


There are a few approaches we could take, as demonstrated by the great solutions by the other tutors.


I'll use the quadratic formula to find the roots. 
Then I'll use those roots to construct the factorization.


The equation {{{2x^2+x-3 = 0}}} is of the form {{{ax^2+bx+c = 0}}}
Notice that: a = 2, b = 1, c = -3


Use the quadratic formula to solve for x.
{{{x = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{x = (-1+-sqrt((1)^2-4(2)(-3)))/(2(2))}}}


{{{x = (-1+-sqrt(1 + 24))/(4)}}}


{{{x = (-1+-sqrt(25))/(4)}}}


{{{x = (-1+-  5)/(4)}}}


{{{x = (-1+5)/(4)}}} or {{{x = (-1-5)/(4)}}}


{{{x = (4)/(4)}}} or  {{{x = (-6)/(4)}}}


{{{x = 1}}} or  {{{x = -3/2}}}
Feel free to skip a few steps if these seem a bit verbose.


The root {{{x = 1}}} leads to {{{x-1 = 0}}} showing that (x-1) is one factor.


The root {{{x = -3/2}}} leads to {{{2x = -3}}} and ultimately {{{2x+3 = 0}}}. 
That makes <font color=red>(2x+3)</font> the other factor.


Ultimately {{{2x^2+x-3}}} factors to {{{(x-1)(2x+3)}}}
You can use the FOIL rule to expand out (x-1)(2x+3) to get the original expression to help verify things. 


We can claim that {{{2x^2+x-3=(x-1)(2x+3)}}} is an identity. 
An identity is a true equation for all valid x in the domain, which in this case is the set of all real numbers.



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Another approach


For this section I'll use the remainder theorem. 
If you prefer the previous section then you can ignore this current section.


The remainder theorem states:
If p(x) divides over (x-k) then p(k) is the remainder.


Consequently an extension is that if (x-k) is a factor of p(x) then p(k) = 0.
The proof is left to the reader.
Hint: p(x) = (x-k)*q(x)


The expression  x+1 is same as writing x - (-1) to show that k = -1 here.
Then that value of k is plugged into x
p(x) = 2x^2 + x - 3
p(-1) = 2(-1)^2 + (-1) - 3
p(-1) = -2
The nonzero result tells us that (x+1) is NOT a factor of 2x^2+x-3


The expression 2x+3 appears to not fit the format x - k.
But with a bit of algebraic adjustments we could say:
2x+3 = 2(x+3/2) = 2(x - (-3/2))


Then we just focus on the x - (-3/2) portion inside.
We have k = -3/2, which leads to,
p(x) = 2x^2 + x - 3
p(-3/2) = 2(-3/2)^2 + (-3/2) - 3
p(-3/2) = 0
The result of 0 tells us that <font color=red>(2x+3) is a factor</font>.


You should find choices C and D produce nonzero results. 
I'll let the student check those.
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